Edexcel P2 2023 January — Question 2 9 marks

Exam BoardEdexcel
ModuleP2 (Pure Mathematics 2)
Year2023
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeShow formula then optimise: cylinder/prism (single variable)
DifficultyModerate -0.3 This is a standard optimization problem with clearly signposted steps: derive a constraint equation, differentiate, find stationary points, verify minimum with second derivative, and calculate the minimum value. All techniques are routine for P2 level with no novel insights required, making it slightly easier than average.
Spec1.02z Models in context: use functions in modelling1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives
  1. In this question you must show all stages of your working.
\section*{Solutions based entirely on calculator technology are not acceptable.} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f6af51c1-5f85-4952-b3c4-9dca42b2a309-04_629_995_411_534} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} A brick is in the shape of a cuboid with width \(x \mathrm {~cm}\) ,length \(3 x \mathrm {~cm}\) and height \(h \mathrm {~cm}\) ,as shown in Figure 2. The volume of the brick is \(972 \mathrm {~cm} ^ { 3 }\)
  1. Show that the surface area of the brick,\(S \mathrm {~cm} ^ { 2 }\) ,is given by $$S = 6 x ^ { 2 } + \frac { 2592 } { x }$$
  2. Find \(\frac { \mathrm { d } S } { \mathrm {~d} x }\)
  3. Hence find the value of \(x\) for which \(S\) is stationary.
  4. Find \(\frac { \mathrm { d } ^ { 2 } S } { \mathrm {~d} x ^ { 2 } }\) and hence show that the value of \(x\) found in part(c)gives the minimum value of \(S\) .
  5. Hence find the minimum surface area of the brick.
Question 2:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((S =)\, 6x^2 + 6xh + 2xh\)B1 Correct expression for surface area in any form
\(V = 3x^2h = 972 \Rightarrow h = \frac{972}{3x^2}\) or \(hx = \frac{324}{x}\); \(\Rightarrow (S=)\, 6x^2 + 8x\!\left(\frac{972}{3x^2}\right)\) or \((S=)\, 6x^2 + 8\!\left(\frac{324}{x}\right)\)M1 Uses volume in dimensionally correct formula to obtain \(h\) or \(hx\) in terms of \(x\); substitutes into \(S\) of form \(...xh+...x^2\)
\(S = 6x^2 + \frac{2592}{x}\) *A1* Achieved with no errors including bracketing omissions; must see separate volume equation, substitution before \(\frac{2592}{x}\) term, and \(S=\) at some point
Part (b)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\left(\frac{dS}{dx} =\right) 12x - \frac{2592}{x^2}\)B1 In any form e.g. \(12x - 2592x^{-2}\)
Part (c)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(12x - \frac{2592}{x^2} = 0 \Rightarrow 12x^3 = 2592\); \(\Rightarrow x = \sqrt[3]{\frac{2592}{12}}\)M1 Starts from derivative of form \(Ax - Bx^{-2}\), sets equal to 0, solves via correct method; \(\sqrt[3]{\frac{2592}{12}}\) scores M1; stating 6 without correct derivative is M0
\(x = 6\)A1 cao; withhold if \(\pm 6\) seen; only possible from correct derivative
Part (d)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\left(\frac{d^2S}{dx^2} =\right) 12 + \frac{5184}{x^3}\)B1ft Correct second derivative; ft their first derivative of form \(Ax - Bx^{-2}\), award \(A + \frac{2B}{x^3}\)
\(\frac{d^2S}{dx^2} > 0\) when \(x=6\) so minimumB1 Fully correct justification with conclusion; must substitute \(x=6\) giving 36 and state \((36)>0\) hence minimum, or justify as \(x>0\) so \(12+\frac{5184}{x^3}>0\); only possible if \(x=6\) found correctly in (c)
Part (e)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(S = 6(6)^2 + \frac{2592}{6} = 648 \text{ (cm}^2)\)B1 \(648\text{ cm}^2\); condone lack of/incorrect units; provided from \(x=6\); \(x=-12\) leading to 648 is B0 
# Question 2:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(S =)\, 6x^2 + 6xh + 2xh$ | B1 | Correct expression for surface area in any form |
| $V = 3x^2h = 972 \Rightarrow h = \frac{972}{3x^2}$ or $hx = \frac{324}{x}$; $\Rightarrow (S=)\, 6x^2 + 8x\!\left(\frac{972}{3x^2}\right)$ or $(S=)\, 6x^2 + 8\!\left(\frac{324}{x}\right)$ | M1 | Uses volume in dimensionally correct formula to obtain $h$ or $hx$ in terms of $x$; substitutes into $S$ of form $...xh+...x^2$ |
| $S = 6x^2 + \frac{2592}{x}$ * | A1* | Achieved with no errors including bracketing omissions; must see separate volume equation, substitution before $\frac{2592}{x}$ term, and $S=$ at some point |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{dS}{dx} =\right) 12x - \frac{2592}{x^2}$ | B1 | In any form e.g. $12x - 2592x^{-2}$ |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $12x - \frac{2592}{x^2} = 0 \Rightarrow 12x^3 = 2592$; $\Rightarrow x = \sqrt[3]{\frac{2592}{12}}$ | M1 | Starts from derivative of form $Ax - Bx^{-2}$, sets equal to 0, solves via correct method; $\sqrt[3]{\frac{2592}{12}}$ scores M1; stating 6 without correct derivative is M0 |
| $x = 6$ | A1 | cao; withhold if $\pm 6$ seen; only possible from correct derivative |

## Part (d)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{d^2S}{dx^2} =\right) 12 + \frac{5184}{x^3}$ | B1ft | Correct second derivative; ft their first derivative of form $Ax - Bx^{-2}$, award $A + \frac{2B}{x^3}$ |
| $\frac{d^2S}{dx^2} > 0$ when $x=6$ so minimum | B1 | Fully correct justification with conclusion; must substitute $x=6$ giving 36 and state $(36)>0$ hence minimum, or justify as $x>0$ so $12+\frac{5184}{x^3}>0$; only possible if $x=6$ found correctly in (c) |

## Part (e)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $S = 6(6)^2 + \frac{2592}{6} = 648 \text{ (cm}^2)$ | B1 | $648\text{ cm}^2$; condone lack of/incorrect units; provided from $x=6$; $x=-12$ leading to 648 is B0 |

---
\begin{enumerate}
  \item In this question you must show all stages of your working.
\end{enumerate}

\section*{Solutions based entirely on calculator technology are not acceptable.}
\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f6af51c1-5f85-4952-b3c4-9dca42b2a309-04_629_995_411_534}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

A brick is in the shape of a cuboid with width $x \mathrm {~cm}$ ,length $3 x \mathrm {~cm}$ and height $h \mathrm {~cm}$ ,as shown in Figure 2.

The volume of the brick is $972 \mathrm {~cm} ^ { 3 }$\\
(a)Show that the surface area of the brick,$S \mathrm {~cm} ^ { 2 }$ ,is given by

$$S = 6 x ^ { 2 } + \frac { 2592 } { x }$$

(b)Find $\frac { \mathrm { d } S } { \mathrm {~d} x }$\\
(c)Hence find the value of $x$ for which $S$ is stationary.\\
(d)Find $\frac { \mathrm { d } ^ { 2 } S } { \mathrm {~d} x ^ { 2 } }$ and hence show that the value of $x$ found in part(c)gives the minimum value of $S$ .\\
(e)Hence find the minimum surface area of the brick.

\hfill \mbox{\textit{Edexcel P2 2023 Q2 [9]}}
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